3.345 \(\int \frac {\cos ^2(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=98 \[ \frac {2 (5 B-2 C) \sin (c+d x)}{3 a^2 d}-\frac {(2 B-C) \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac {x (2 B-C)}{a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-(2*B-C)*x/a^2+2/3*(5*B-2*C)*sin(d*x+c)/a^2/d-(2*B-C)*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(B-C)*sin(d*x+c)/d/(
a+a*sec(d*x+c))^2

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Rubi [A]  time = 0.32, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4072, 4020, 3787, 2637, 8} \[ \frac {2 (5 B-2 C) \sin (c+d x)}{3 a^2 d}-\frac {(2 B-C) \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac {x (2 B-C)}{a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x]

[Out]

-(((2*B - C)*x)/a^2) + (2*(5*B - 2*C)*Sin[c + d*x])/(3*a^2*d) - ((2*B - C)*Sin[c + d*x])/(a^2*d*(1 + Sec[c + d
*x])) - ((B - C)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cos (c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\\ &=-\frac {(B-C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\cos (c+d x) (a (4 B-C)-2 a (B-C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {(2 B-C) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(B-C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \cos (c+d x) \left (2 a^2 (5 B-2 C)-3 a^2 (2 B-C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {(2 B-C) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(B-C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(2 (5 B-2 C)) \int \cos (c+d x) \, dx}{3 a^2}-\frac {(2 B-C) \int 1 \, dx}{a^2}\\ &=-\frac {(2 B-C) x}{a^2}+\frac {2 (5 B-2 C) \sin (c+d x)}{3 a^2 d}-\frac {(2 B-C) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(B-C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}

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Mathematica [B]  time = 0.65, size = 245, normalized size = 2.50 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-18 d x (2 B-C) \cos \left (c+\frac {d x}{2}\right )-30 B \sin \left (c+\frac {d x}{2}\right )+41 B \sin \left (c+\frac {3 d x}{2}\right )+9 B \sin \left (2 c+\frac {3 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {5 d x}{2}\right )-12 B d x \cos \left (c+\frac {3 d x}{2}\right )-12 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-18 d x (2 B-C) \cos \left (\frac {d x}{2}\right )+66 B \sin \left (\frac {d x}{2}\right )+24 C \sin \left (c+\frac {d x}{2}\right )-20 C \sin \left (c+\frac {3 d x}{2}\right )+6 C d x \cos \left (c+\frac {3 d x}{2}\right )+6 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-36 C \sin \left (\frac {d x}{2}\right )\right )}{12 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-18*(2*B - C)*d*x*Cos[(d*x)/2] - 18*(2*B - C)*d*x*Cos[c + (d*x)/2] - 12*B*d*x*Cos[
c + (3*d*x)/2] + 6*C*d*x*Cos[c + (3*d*x)/2] - 12*B*d*x*Cos[2*c + (3*d*x)/2] + 6*C*d*x*Cos[2*c + (3*d*x)/2] + 6
6*B*Sin[(d*x)/2] - 36*C*Sin[(d*x)/2] - 30*B*Sin[c + (d*x)/2] + 24*C*Sin[c + (d*x)/2] + 41*B*Sin[c + (3*d*x)/2]
- 20*C*Sin[c + (3*d*x)/2] + 9*B*Sin[2*c + (3*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2] + 3*B*Sin[3*c + (5*d*x)/2]))/
(12*a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 0.49, size = 123, normalized size = 1.26 \[ -\frac {3 \, {\left (2 \, B - C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (2 \, B - C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (2 \, B - C\right )} d x - {\left (3 \, B \cos \left (d x + c\right )^{2} + {\left (14 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 10 \, B - 4 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(3*(2*B - C)*d*x*cos(d*x + c)^2 + 6*(2*B - C)*d*x*cos(d*x + c) + 3*(2*B - C)*d*x - (3*B*cos(d*x + c)^2 +
(14*B - 5*C)*cos(d*x + c) + 10*B - 4*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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giac [A]  time = 0.32, size = 121, normalized size = 1.23 \[ -\frac {\frac {6 \, {\left (d x + c\right )} {\left (2 \, B - C\right )}}{a^{2}} - \frac {12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} + \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(6*(d*x + c)*(2*B - C)/a^2 - 12*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) + (B*a^4*tan(1/
2*d*x + 1/2*c)^3 - C*a^4*tan(1/2*d*x + 1/2*c)^3 - 15*B*a^4*tan(1/2*d*x + 1/2*c) + 9*C*a^4*tan(1/2*d*x + 1/2*c)
)/a^6)/d

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maple [A]  time = 1.10, size = 149, normalized size = 1.52 \[ -\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {3 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)

[Out]

-1/6/d/a^2*B*tan(1/2*d*x+1/2*c)^3+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+5/2/d/a^2*B*tan(1/2*d*x+1/2*c)-3/2/d/a^2*C*
tan(1/2*d*x+1/2*c)+2/d/a^2*B*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-4/d/a^2*arctan(tan(1/2*d*x+1/2*c))*B+
2/d/a^2*arctan(tan(1/2*d*x+1/2*c))*C

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maxima [B]  time = 0.44, size = 191, normalized size = 1.95 \[ \frac {B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - C {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(B*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1
))) - C*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2))/d

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mupad [B]  time = 2.81, size = 109, normalized size = 1.11 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B-C}{a^2}+\frac {3\,B-C}{2\,a^2}\right )}{d}-\frac {x\,\left (2\,B-C\right )}{a^2}+\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^2,x)

[Out]

(tan(c/2 + (d*x)/2)*((B - C)/a^2 + (3*B - C)/(2*a^2)))/d - (x*(2*B - C))/a^2 + (2*B*tan(c/2 + (d*x)/2))/(d*(a^
2*tan(c/2 + (d*x)/2)^2 + a^2)) - (tan(c/2 + (d*x)/2)^3*(B - C))/(6*a^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(B*cos(c + d*x)**2*sec(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*cos(c + d*x)*
*2*sec(c + d*x)**2/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

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